1. ## Integral of secant

Ok- so....why is this not possible?

∫sec(θ)= ∫1/cos(θ)= ln(cosθ)*(-sinθ)

2. ## Re: Integral of secant

Hey CoffeeBird.

If we take the derivative of ln(cos(theta))*-sin(theta) then we should get 1/cos(theta) by the fundamental theorem of calculus (which relates anti-derivatives to derivatives).

So d/d(theta) (ln(cos(theta))*-sin(theta) =
= vu' + uv' = -sin(theta)*sin(theta)/cos(theta) + ln(cos(theta))*-cos(theta)
= -sin(theta)tan(theta) - cos(theta)*ln(cos(theta)).

Now google says that the derivative is ln|sec(theta) + tan(theta)|. Lets differentiate this and see what we get:

d/d(theta) ln|sec(theta) + tan(theta)| = d/d(theta)sec(theta) + d/d(theta)tan(theta) / [sec(theta) + tan(theta)]
= [tan(theta)sec(theta) + sec^2(theta)]/[sec(theta) + tan[theta])
= sec(theta)[tan(theta) + sec(theta)]/[tan(theta) + sec(theta)]
= sec(theta) * 1
= sec(theta).

3. ## Re: Integral of secant

Originally Posted by CoffeeBird
Ok- so....why is this not possible?

∫sec(θ)= ∫1/cos(θ)= ln(cosθ)*(-sinθ)
\displaystyle \begin{align*} \int{\sec{x}\,dx} &= \int{\frac{1}{\cos{x}}\,dx} \\ &= \int{\frac{\cos{x}}{\cos^2{x}}\,dx} \\ &= \int{\frac{\cos{x}}{1 - \sin^2{x}}\,dx} \\ &= \int{\frac{1}{1 - u^2}\,du} \textrm{ after making the substitution } u = \sin{x} \implies du = \cos{x}\,dx \\ &= \int{\frac{1}{(1 - u)(1 + u)}\,du} \\ &= \int{\frac{1}{2(1 - u)} + \frac{1}{2(1 + u)}\,du} \\ &= -\frac{1}{2}\ln{|1 - u|} + \frac{1}{2}\ln{|1 + u|} + C \\ &= \frac{1}{2}\ln{\left| \frac{1 + u}{1 - u} \right|} + C \\ &= \frac{1}{2}\ln{\left| \frac{1 + \sin{x}}{1 - \sin{x}} \right|} + C \\ &= \frac{1}{2}\ln{\left| \frac{(1 + \sin{x})^2}{(1 - \sin{x})(1 + \sin{x})} \right|} + C \\ &= \frac{1}{2}\ln{\left| \frac{(1 + \sin{x})^2}{1 - \sin^2{x}} \right|} + C \\ &= \frac{1}{2}\ln{\left| \frac{(1 + \sin{x})^2}{\cos^2{x}} \right|} + C \\ &= \ln{\left| \left[ \frac{(1 + \sin{x})^2}{\cos^2{x}} \right]^{\frac{1}{2}} \right|} + C \end{align*}

\displaystyle \begin{align*} &= \ln{\left| \frac{1 + \sin{x}}{\cos{x}} \right|} + C \\ &= \ln{\left| \frac{1}{\cos{x}} + \frac{\sin{x}}{\cos{x}} \right|} + C \\ &= \ln{\left| \sec{x} + \tan{x} \right|} + C \end{align*}

4. ## Re: Integral of secant

Another approach:

$\int\sec$$\theta$$\,d\theta$

Let:

$u=\sec(\theta)+\tan(\theta)\:\therefore\:du=(\sec( \theta)\tan(\theta)+\sec^2(\theta))\,d\theta=u\sec (\theta)\,d\theta$

giving:

$\int\frac{du}{u}=\ln|u|+C=\ln|\sec(\theta)+\tan( \theta)|+C$

5. ## Re: Integral of secant

Originally Posted by CoffeeBird
Ok- so....why is this not possible?

∫sec(θ)= ∫1/cos(θ)= ln(cosθ)*(-sinθ)
Because the chain rule process...

... is in the direction of differentiation (i.e. downwards), whether you're differentiating...

... which works perfectly well for differentiating sec...

... or whether you are instead anti-differentiating. In which case, you are definitely NOT trying to do this...

... but rather this...

Hence chiro and MarkFL2's suggestion...

... where chiro is checking the differentiation going clockwise from top left, and MarkFL2 the reverse, starting bottom left.

Another way (more roundabout but more generally applicable to trigs in fractions) is...

See here for a step by step guide for the very similar cosec integral by this method.

And ProveIt's would make a nice picture, too... try it?!

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